CBSE Sample Papers for Class 12 Physics Term 2 Set 8 with Solutions

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CBSE Sample Papers for Class 12 Physics Standard Term 2 Set 8 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 40

General Instructions:

• There are 12 questions in all All questions are compulsory.
• This question paper has three sections: Section A, Section B and Section C.
• Section A contains three questions of two marks each, Section B contains eight questions of three
  marks each, Section C contains one case study-based question of five marks.
• There is no overall choice However, an internal choice has been provided en one question of two marks and two questions of three marks You have to attempt only one of the choices in such questions
• You may use log tables if necessary but use of calculator is not allowed.

Section – A

Question 1.
In a pure semiconductor crystal of Si, if antimony is added then what type of extrinsic semiconductor is obtained. Draw the energy band diagram of this extrinsic semiconductor so formed.
Answer:
As antimony is added to pure Si crystal, then a K-type extrinsic semiconductor would be obtained, since antimony (Sb) is a pentavalent impurity.
Energy level diagram of n-type semiconductor:

Question 2.
Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but same orbital angular momentum according to the Bohr model? Justify your answer.
Answer:
No, the electrons with different energies cannot have the same angular momentum. This is because
according to Bohr model, E_n = \(\frac{13.6}{n^{2}}\)eV. When E_n is different, n must be different. And angular momentum, mυr = \(\frac{n h}{2 \pi}\) must be different for different n values.
OR
Explain how does (i) photoelectric current and (ii) kinetic energy of the photoelectrons emitted in a photocell vary if the frequency of incident radiation is doubled, but keeping the intensity same?
Show the graphical variation in the above two cases.
Answer:
(i) The increase in the frequency of incident radiation has no effect on photoelectric current. This is because of incident photon of increased energy cannot eject more than one electron from the metal surface.

(ii) The kinetic energy of photoelectrons becomes more than the double of its original energy. As the work function of the metal is fixed, so incident photon of higher frequency and hence higher energy will impart more energy to the photoelectrons.

Question 3.
Name the device which converts the change in intensity of illumination to change in electric current flowing through it. Plot I-V characteristics of this device for different intensities. State any two applications of this device.
Answer:
Photodiodes are used to detect optical signals of different intensities by changing current flowing through them.
I-V characteristics of a Photodiode:

Applications of Photodiodes:

1. In detection of optical signals.
2. In demodulation of optical signals.
3. In light operated switches.
4. In speed reading of computer punched cards.
5. In electric counters.

SECTION B

Question 4.
Derive an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n – 1). Also show that for large values of n, this frequency equals to classical frequency of revolution of an electron.
Answer:
From Bohr’s theory, the frequency/of the radiation emitted when an electron de-excites from level n₂ to level n₁ is given as:
f = \(\frac{2 \pi^{2} m k^{2} z^{2} e^{4}}{h^{2}}\) [latex]\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}[/latex]
Given n₁ = n – 1,
n₂ = n,
f = \(\frac{2 \pi^{2} m k^{2} z^{2} e^{4}}{h^{2}} \frac{(2 n-1)}{(n-1)^{2} n^{2}}\)
For large values of n, 2n – 1 ≈ 2 n,
n – 1 ≈ n
and z = 1
Thus, f = \(\frac{4 \pi^{2} m k^{2} e^{4}}{n^{3} h^{3}}\)
Which is same as orbital frequency of electron in n^th orbit.
f = \(\frac{v}{2 \pi r}=\frac{4 \pi^{2} m k^{2} e^{4}}{n^{3} h^{3}}\)

Question 5.
Explain with a proper diagram how an ac signal can be converted into dc (pulsating) signal with output frequency as double than the input frequency using p-n junction diode. Give its input and output waveforms.
Answer:
A junction diode allows current to pass only when it is forward biased. So, if an alternating voltage is applied across a diode the current flows only in that part of the cycle when the diode is forward biased. This property is used to rectify alternating voltages and the circuit used for this purpose is called a rectifier.
Circuit Diagram:

Question 6.
How long can an electric lamp of 100 W be kept glowing by fusion of 2 kg of deuterium? Take the fusion reaction as:
\({ }_{1}^{2} \mathrm{H}\) + \({ }_{1}^{2} \mathrm{H}\) → \({ }_{2}^{3} \mathrm{He}\) + n + 3.27 MeV
Answer:
Number of atoms present in 2 g of deuterium = 6.023 × 10²³
Number of atoms present m 2 kg of deuterium = \(\frac{6.023 \times 10^{23} \times 2000}{2}\) = 6.023 × 10²⁶
Energy released in fusion of 2 deuterium atoms = 3.27 MeV
Energy released in fusion of 2 kg of deuterium atoms
= \(\frac{3.27}{2}\) × 6.023 × 10²⁶ MeV
= 9.85 × 10²⁶ MeV
= 15.75 × 10¹³ J
Energy consumed by bulb per sec = 100 J
Time for which bulb will glow = \(\frac{15.75 \times 10^{13}}{100}\) s = 15.75 × 10¹¹sec
Number of years = \(\frac{15.75 \times 10_{11} \mathrm{sec}}{365 \times 24 \times 60 \times 60}\)
= 4.99 × 104 years.

Question 7.
Define wavefront. Draw the shape of refracted wavefront when the plane incident wave undergoes refraction from optically denser medium to rarer medium. Hence prove Snell’s law of refraction.
Answer:
(i) Wavefront: The continuous locus of all the particles of a medium, which are vibrating in the same phase is called a wavefront.

(ii) Snell’s law of refraction: Let PP’ represent the surface separating medium 1 and medium 2 as shown in figure.
Let υ₁ and υ₂ represent the speed of light in medium 1 and medium 2 respectively. We assume a plane wavefront AB propagating in the direction A’A incident on the interface at an angle i. Let t be the time taken by the wavefront to travel the distance BC.
∴ BC = υ₁ [distance = speed × time]
In order to determine the shape of the refracted wavefront, we draw a sphere of radius υ₂t from the point A in the second medium (the speed of the wave in second medium is υ₂)
Let CE represent a tangent plane drawn from the point C. Then AE = υ₂t.
∴ CE would represent the refracted wavefront.

In ΔABC and ΔAEC, we have
sin i = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{v_{1} t}{\mathrm{AC}}\) and sin r = \(\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{v_{2} t}{\mathrm{AC}}\)
where i and r are angles of incident and refraction respectively.
∴ \(\frac{\sin i}{\sin r}\) = \(\frac{v_{1} t}{\mathrm{AC}} \cdot \frac{\mathrm{AC}}{v_{2} t}\)
⇒ \(\frac{\sin i}{\sin r}\) = \(\frac{v_{1}}{v_{2}}\)
If c represents the speed of light jn vacuum, then
n₁ = \(\frac{c}{v_{1}}\) and n₂ = \(\frac{c}{v_{2}}\)
⇒ υ₁ = \(\frac{\mathcal{c}}{n_{1}}\) and υ₂ = \(\frac{\mathcal{c}}{n_{2}}\)
Where n₁ and n₂ are the refractive indices of medium 1 and medium 2.
∴ \(\frac{\sin i}{\sin r}=\frac{c / n_{1}}{c / n_{2}}\)
⇒ \(\frac{\sin i}{\sin r}=\frac{n_{2}}{n_{1}}\)
⇒ n₁ sin i = n₂ sin r
This is the Snell’s law of refraction.

Question 8.
(a) Draw a ray diagram of compound microscope for the final image formed at least distance of distinct vision?
(b) An angular magnification of 30X is desired using an objective of focal length 1.25 cm and an eye piece of focal length 5 cm. How will you set up the compound microscope for the final image formed at least distance of distinct vision?
Answer:
(a) Ray diagram of a compound microscope for the final image formed at least distance of distinct vision:

(b) Given: m = 30, f₀ = 1.25 cm, f_e = 5 cm
When image is formed at least distance of distinct vision,
D = 25 cm
Angular magnification of eyepiece
m_e = (1 + \(\frac{\mathrm{D}}{f_{e}}\)) = 1 + \(\frac{25}{5}\) = 6
Total Angular magnification, m = m₀m_e
⇒ m₀ = \(\frac{m}{m_{e}}\) = \(\frac{30}{6}\) = 5
As the objective lens forms the real image.
m₀ = \(\frac{v_{0}}{u_{0}}\) = -5
⇒ υ₀ = -5u₀
Using lens equation, we get, usub>0 = -15 cm, υ₀ = -5 × (-1.5) cm = + 7.5 cm
Given υ_e = -D = -25 cm , f_e = + 5 cm, u_e = ?
Using again lens equation, we get, u_e = \(\frac{25}{6}\)
Thus, object is to be placed at 1.5 cm from the objective and separation between the two lenses should be
L = υ₀ + Iu_eI = 11.67 cm.

OR

(a) Draw a ray diagram of Astronomical Telescope for the final image formed at infinity.
(b) A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. Find the magnifying power of the telescope for viewing distant objects when:
(i) the telescope is in normal adjustment
(ii) the final image is formed at the least distance of distinct vision.
Answer:
(a) Ray diagram of a Astronomical telescope for the final image formed as infinity:

(b) (i) Given:
Focal length of the objective lens, f₀ = 140 cm
Focal length of the eyepiece, f_e = 5 cm
Least distance of distinct vision, D = 25 cm
When the telescope is in normal, adjustment, its magnifying power is given as:
m = –\(\frac{f_{0}}{f_{e}}\) = -28
(ii) When the final image is formed at d, its magnifying power is given as:
m = – \(\frac{f_{0}}{f_{e}}\)(1 + \(\frac{f_{e}}{\mathrm{D}}\))
= \(\frac{140}{5}\)(1 + \(\frac{5}{25}\)) = -33.6

Question 9.
Light of wavelength 2000 Å falls on a metal surface of work function 4.2 eV.
(a) What is the kinetic energy (in eV) of the fastest electrons emitted from the surface?
(b) What will be the change in the energy of the emitted electrons if the intensity of light with same , wavelength is doubled?
(c) If the same light falls on another surface of work function 6.5 eV, what will be the energy of emitted electrons?
Answer:
(a) λ = 2000 Å = 2000 × 10^-10 m
W₀ = 4.2 eV
h = 6.63 × 10^-34
\(\frac{h c}{\lambda}\) = W₀ + KE
or K.E = \(\frac{h c}{\lambda}\) – W₀
= \(\frac{\left(6.63 \times 10^{-34}\right) \times\left(3 \times 10^{8}\right)}{\left(2000 \times 10^{-10}\right)}\) × \(\frac{1}{1.6 \times 10^{-19}}\) eV – 4.2 eV
= (6.2 – 4.2) eV = 2.0 eV.

(b) The energy of the emitted electrons does not depend upon intensity of incident light; hence the energy remains unchanged.

(c) For this surface, electrons will not be emitted as the energy of incident light (6.2 eV) is less than the work function (6.5 eV) of the surface.

Question 10.
The focal length of a convex lens made of glass of refractive index (1.5) is 20 cm.
What will be its new focal length when placed in a medium of refractive index 1.25 ?
Is focal length positive or negative? What does it signify?
Answer:
Given, a_μg = 1.5
Focal length of the given convex lens which it is placed in air is
f = + 20 cm
\(\frac{1}{f}\) = (a_μg – 1) [(\(\frac{1}{R_{1}}\)) – (\(\frac{1}{R_{2}}\))] ……….(A)
Refractive index of the given medium with respect to air is
a_μm = 1.25
New focal length of the given convex lens when placed in a medium is f’.
\(\frac{1}{f^{\prime}}\) = (m_μg – 1) [(\(\frac{1}{R_{1}}\)) – (\(\frac{1}{R_{2}}\))] …….. (B)
Dividing (A) by (B) we get
\(\frac{f^{\prime}}{f}=\frac{\left(a_{\mu_{g}}-1\right)}{\left(m_{\mu_{g}}-1\right)}\) = \(\frac{(1.5-1)}{(1.2-1)}=\frac{0.5}{0.2}=\frac{5}{2}\) = 2.5
f’ = 2.5 f = (2.5 × 20)cm = + 50 cm as m_μg = \(\) = \(\) = 1.2
New focal length is positive.
The significance of the positive sign of the focal length is that given convex lens is still converging in the given medium.

Question 11.
(a) Name the E.M. waves which are suitable for radar systems used in aircraft navigation. Write the range of frequency of these waves.
(b) If the Earth did not have atmosphere, would its average surface temperature be higher or lower than what it is now? Explain.
(c) An E.M. wave exerts pressure on the surface on which it is incident. Justify.
Answer:
(a) Microwaves are suitable for radar systems used in aircraft navigation. The range of frequency for i these waves is 10⁹ Hz to 10¹²Hz.

(b) There would be no greenhouse effect on the surface of the Earth in the absence of atmosphere. As a result, the temperature of the Earth would decrease rapidly, making it difficult for human survival.

(c) When the wave is incident on the metal surface, it is completely absorbed. Energy U and hence momentum (p = E/C) is delivered to the surface of the earth. The momentum delivered becomes twice when, the wave is totally reflected because momentum is charged from pto-p. Thus, force is acting and thereby pressure is exerted on the surface of the earth by EM waves.

OR
(a) “If the slits in Young’s double slit experiment are identical, then intensity at any point on the screen may vary between zero and four times to the intensity due to single slit”.
Justify the above statement through a relevant mathematical expression.
(b) Draw the intensity distribution as function of phase angle when diffraction of light takes place through co-herently illuminated single slit.
Answer:
(a) The total intensity at a point where the phase difference is Φ, is given by I = I₁ + I₂ + 2\(\sqrt{\mathrm{I}_{1} \mathrm{I}_{2}}\) cos Φ.
Here I₁ and I₂ are the intensities of two individual sources which are equal i.e., I₁ = I₂.
When Φ is 0,I = 4I₁
When Φ is 180°, I = 0.
Thus intensity on the screen varies between 4I₁ and 0.

(b) Intensity distribution as function of phase angle, when diffraction of light takes place through co-herently illuminated single slit.
The intensity pattern on the screen is shown in the given figure.

Width of central maximum = \(\frac{2 \mathrm{D} \lambda}{a}\)

SECTION C

Question 12.
CASE STUDY: MIRAGE IN DESERTS

To a distant observer, the light appears to be coming from somewhere below the ground. The observer naturally assumes that light is being reflected from the ground, say, by a pool water near the tall object.

Such inverted images of distant tall objects cause an optical illusion to the observer. This phenomenon is called mirage. This type of mirage is especially common in hot deserts.

Based on the above facts, answer the following questions:
(a) Which of the following phenomena is prominently involved in the formation of mirage in deserts? [1]
(i) Refraction, Total internal Reflection
(ii) Dispersion and Refraction
(iii) Dispersion and scattering of light
(iv) Total internal Reflection and diffraction

(b) A diver at a depth 12 m inside water (a_μω = \(\frac{4}{3}\))sees the sky in a cone of semi- vertical angle: [1]
(i) sin^-1\(\frac{4}{3}\)
(ii) tan^-1 \(\frac{4}{3}\)
(iii) sin^-1\(\frac{3}{4}\)
(iv) 90°

(c) In an optical fibre, if n₁ and n₂ are the refractive indices of the core and cladding, then which among the following, would be a correct equation? [1]
(i) n₁ < n₂
(ii) n₁ = n₂
(iii) n₁ << n₂
(iv) n₁> n₂

(d) A diamond is immersed in such a liquid which has its refractive index with respect to air as greater than the refractive index of water with respect to air. Then the critical angle of diamond-liquid interface as compared to critical angle of diamond -water interface will: [1]
(i) depend on the nature of the liquid only
(iii) remain the same
(ii) decrease
(iv) increase.

(e) The following figure shows a cross-section of a Tight pipe’ made of a glass fiber of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for the following phenomena to
occur? [1]

(i) 0 < i < 90°
(ii) 0 < i < 60°
(iii) 0 < i < 45°
(iv) 0 < i <30°
Answer:
(a)
(i) Refraction, Total internal reflection
Explanation: The mirage is formed because of Total Internal Reflection and refraction.
Mirage is a naturally occurring optical illusion in which light rays bend to produce a displaced image of distant objects or the sky.

(b) (iii) sin^-1\(\frac{3}{4}\)
Explanation: Given μ = \(\frac{4}{3}\) and μ_air = 1
The ray emerges out of water making an angle 90° with the water surface, i.e., Using Snell’s law of refraction at point A:
im
⇒ μ sin θ = μ_air sin r
⇒ \(\frac{4}{3}\) × sin θ = 1 × sin 90°
⇒ \(\frac{4}{3}\) × sin θ = 1 × 1
⇒ sinθ = \(\frac{3}{4}\)
⇒ θ = sin^-1(\(\))

(c) (iv) n₁ > n₂
Explanation: The refractive index of the core should be greater than the refractive index of the cladding.

(d) (iv) Increases
Explanation: l_μd = \(\frac{1}{\sin C}=\frac{\mu_{d}}{\mu_{l}}\), ω_μd = \(\frac{1}{\sin C^{\prime}}=\frac{\mu_{d}}{\mu_{\omega}}\)
μ_l > μ_ω
Thus C > C’.

(e) (ii) 0 < i < 60°
Explanation: 1_μ2 = \(\frac{1}{\sin C^{\prime}}\)
sin C’ = \(\frac{1.44}{1.68}\) = 0.8571
⇒ C = 59°
Total internal reflection will occur if the angle i’ > i_c‘.
i.e., if i’ > 59° or when r < r_max where r_max = 90° – 59° = 31°.
Using Snell’s law,
\(\frac{\sin i_{\max }}{\sin r_{\max }}\) = 1.68
or sin i_max = 1.68 × sin r_max
= 1.68 × sin 31°
= 1.68 × 0.5150
= 0.8662
∴ i_max = 60°
Thus all incident rays which makes angles in the range 0 < i
60° with the axis of the pipe will suffer total internal reflections in the pipe.